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Proof step 10 from proof eq (previous step)

a0 ' + ins≤(a, x0) = ins≤(a, y), a0 < a, a0 ∈ ins≤(a, y), ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 ' + x0 = y

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ x ≠ [] ∧ x.head < a → ins≤(a, x) = x.head + ins≤(a, x.tail)
Proof step 12:
a0 ' + ins≤(a, x0) = ins≤(a, y), a0 < a, a0 ∈ ins≤(a, y), ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 ' + x0 = y, y ≠ [] ∧ y.head < a
Proof step 14:
a0 ' + ins≤(a, x0) = y.head + ins≤(a, y.tail), a0 < a, y.head < a, a0 ∈ y.head + ins≤(a, y.tail), ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 ' + x0 = y, y = []

Rule argument

spec: <toplevel> name: ins-rec seq: ⊦ x ≠ [] ∧ x.head < a → ins≤(a, x) = x.head + ins≤(a, x.tail) subst: [x, a] → [y, a] flags: <all paths>

Additional information

Used terms: y, a